Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes [tex]P_B(A)[/tex].
The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
[tex]P(A|B)= \frac{P(A \cap B)}{P(B)} [/tex]
In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.
The probability of selecting one coin is [tex] \frac{1}{3} [/tex]
Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.
P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.
Thus [tex]P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27} [/tex]
P(B) means that the first envelope contains a quarter AND the
second envelope contains a quarter
Thus [tex]P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} [/tex]
Therefore, [tex]P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3} [/tex]
Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
[tex]P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9} [/tex]
[tex]P(D)= \frac{1}{3} [/tex]
Therefore, [tex]P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3} [/tex]
